# Pairing of cotangent and tangent bundles

I am reading the survey paper: "The de-Rham Witt complex and Crystalline cohomology" by Luc Illusie.

In math line (2.1.12), Illusie considers the pairing $$\langle-,-\rangle:\Omega_{X/S}^1\times T_{X/S}\longrightarrow \mathcal{O}_X$$ of the tangent and cotangent bundles on a scheme $$X$$ of characteristic $$p$$, relative to a morphism $$X\longrightarrow S$$. ($$S$$ also has characteristic $$p$$).

Earlier we defined a bunch of maps, which would be necessary to describe in order for me to ask my question. The first one is the Cartier operation, $$C$$, which sends a closed form $$w\in \Omega_{X/S}^1$$ to a 1-form $$Cw \in \Omega_{X^{(p)}/S}^1$$, where $$X^{(p)}$$ is the base change of $$X\longrightarrow S$$ with respect to the absolute Frobenius on $$S$$. Denote by $$W$$ the canonical projection $$W:X^{(p)}\longrightarrow X$$.

Illusie then wants to say something about the pairing $$\langle Cw,W^*D\rangle$$ which happens on $$\Omega_{X^{(p)}/S}^1\times T_{X^{(p)}/S}\longrightarrow \mathcal{O}_{X^{(p)}}$$. Namely, he gives the identity $$\langle Cw,W^*D\rangle^p = \langle w,D^p\rangle - D^{p-1}\langle w,D\rangle.$$ My questions are:

1. What does the $$p$$-th power of a tangent vector, $$D^p$$, mean? Does the tangent space have a ring structure?

2. The pairing apriori should have values in $$\mathcal{O}_{X^{(p)}}$$, once it is raised to the $$p$$'th power we regard it as having values in $$\mathcal{O}_X$$ as this is. For this reason it seems that Illusie is regarding $$D^{p-1}$$ as an element of $$\mathcal{O}_X$$, which is consistent with math line (2.1.13), why is that? I guess this goes back to the first question.

3. Illusie also writes $$D_i^p = 0$$, where $$D_i = \partial/\partial x_i$$, for some etale basis $$(x_i)$$, why is that true? I guess that this should all be clear once I understand this notation.

• Think of $D$ as a derivation, a kind an operator on functions. The composition of derivations is not generally a derivation, but the iterated operator $D^p$ is a derivation if you are in characteristic $p$. Also $D_i^p=0$ in characteristic $p$ (the $p$th derivative of a polynomial with respect to one of its variables has a factor of $p$ factorial). Aug 1, 2022 at 18:30
• Thanks, you definitely noticed a gap in my understanding! I didn't think about $D$ as being an operator on the coordinate ring, but rather as an element of the co-cotangent bundle. Aug 1, 2022 at 22:24

For (1), recall that if $$R$$ is a ring, then a derivation $$D: R \to R$$ satisfies the Leibniz rule, which by induction on $$n$$ implies that if $$D^n$$ denotes the $$n$$-fold iterate of $$D$$, then $$D^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i(f) D^{n-i}(g).$$ Since $$\binom{p}{i} \equiv 0$$ for $$0, this implies that if $$R$$ is an $$\mathbf{F}_p$$-algebra, then $$D^p(fg) = D^p(f) g + f D^p(g)$$. In other words, $$D^p$$ is a derivation.

Let me answer (3) before (2). If you work locally, i.e., consider the derivation $$\partial_x$$ on $$\mathbf{F}_p[x]$$, then $$(\partial_x)^p x^n$$ is zero for $$n, and is $$p! \binom{n}{p} x^{n-p}$$ for $$n\geq p$$, which is zero. So the derivation $$(\partial_x)^p$$ is identically zero.

Let's now discuss (2); since everything is local on $$X$$ and $$X$$ is smooth, we can assume that $$X$$ is etale over $$\mathbf{A}^n_S$$ with basis $$x_1, \cdots, x_n$$. Let $$\omega$$ be a closed $$1$$-form on $$X$$. Because of the Cartier isomorphism $$\mathfrak{C}: \mathcal{H}^i(F_\ast \Omega^\bullet_{X/S}) \xrightarrow{\sim} \Omega^i_{X^{(p)}/S}$$, we can write $$\omega = df + \sum_{i=1}^n F^\ast(g_i) x_i^{p-1} dx_i$$ for some functions $$g_i$$ and $$f$$ on $$X$$. Both $$\langle \mathfrak{C} \omega, D\rangle^p$$ and $$\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle$$ kill $$df$$, so by Frobenius semilinearity, we can assume that $$\omega = x_i^{p-1} dx_i$$. We'll also just assume $$n=1$$ and write $$x$$ instead of $$x_1$$.

Then $$\langle \mathfrak{C} \omega, D\rangle = \langle dx^{(p)}, D\rangle = D(x)$$, and $$\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx)$$. So we need to prove that $$(Dx)^p = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx),$$ which is an identity due to Hochschild. See https://joshuamundinger.github.io/assets/notes/hochschild-identity.pdf for a cute argument; it reduces to using the multinomial analogue of the Leibniz rule for $$D^p(x^p)$$ and a multinomial analogue of the binomial coefficient vanishing.

Clarification from comments: let's again just consider a single variable $$x$$ and assume $$X = \mathbf{A}^1_{\mathbf{F}_p}$$. The most general closed $$1$$-form is $$\omega = df + g(x^p) x^{p-1} dx$$. Then $$\mathfrak{C}(df) = 0$$, so we can take $$a_i$$ to be $$g(x^p) x^{p-1}$$. Now, $$\partial_x$$ is $$\mathbf{F}_p[x^p]$$-linear, so $$\partial_x^{p-1} (g(x^p) x^{p-1}) = g(x^p) \cdot (\partial_x^{p-1} x^{p-1}) = (p-1)! g(x^p) = -g(x^p).$$ This means that $$\mathfrak{C}(g(x^p) x^{p-1} dx) = g(x^{(p)}) dx^{(p)}$$ can be written as $$-\partial_x^{p-1} (g(x^p) x^{p-1}) dx^{(p)}$$, as desired.

• Brilliant answer, thank you so much! You've made this extremely clear, that's amazing. Aug 1, 2022 at 22:24
• Maybe one additional small question: do you know why Illusie writes, in math line 2.1.13, that if $w = \sum a_idx_i$, and $Cw = \sum c_iW^*(dx_i)$, where the $c_i\in \mathcal{O}_{X^{(p)}}$, then $F_{X/S}(c_i) = -(D_i)^{p-1}a_i$? After all, in degree 0, $C^{-1}_{X/S}$ is defined by $F_{X/S}$, so that if $C$ is inverse to it, I would imagine that $F_{X/S}(c_i) = a_i$ instead. This doesn't interrupt my understanding of the whole Lemma, but seems strange. Aug 2, 2022 at 11:21
• Edited to clarify @kindasorta
– skd
Aug 2, 2022 at 13:45